Question: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $16.4$ years; the standard deviation is $3.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living less than $20.2$ years.
Explanation: $16.4$ $12.6$ $20.2$ $8.8$ $24$ $5$ $27.8$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $16.4$ years. We know the standard deviation is $3.8$ years, so one standard deviation below the mean is $12.6$ years and one standard deviation above the mean is $20.2$ years. Two standard deviations below the mean is $8.8$ years and two standard deviations above the mean is $24$ years. Three standard deviations below the mean is $5$ years and three standard deviations above the mean is $27.8$ years. We are interested in the probability of a porcupine living less than $20.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the porcupines will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $12.6$ years and the other half $({16\%})$ will live longer than $20.2$ years. The probability of a particular porcupine living less than $20.2$ years is ${68\%} + {16\%}$, or $84\%$.